Certainly! Let's go through the step-by-step solution for the given system of linear equations:
[tex]\[
\left\{
\begin{array}{l}
4x + y = 21 \\
3x - y = 14
\end{array}
\right.
\][/tex]
1. Label the equations for reference:
[tex]\[
\begin{aligned}
(1) \quad & 4x + y = 21 \\
(2) \quad & 3x - y = 14
\end{aligned}
\][/tex]
2. Add the equations to eliminate [tex]\( y \)[/tex]:
\begin{align}
(1) & \quad 4x + y = 21 \\
(2) & \quad 3x - y = 14 \\
\end{align}
Adding the two equations:
[tex]\[
(4x + y) + (3x - y) = 21 + 14
\][/tex]
Simplify by combining like terms:
[tex]\[
7x = 35
\][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{35}{7} = 5
\][/tex]
4. Substitute [tex]\( x = 5 \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]:
Using equation (1):
[tex]\[
4x + y = 21
\][/tex]
Substitute [tex]\( x = 5 \)[/tex]:
[tex]\[
4(5) + y = 21
\][/tex]
[tex]\[
20 + y = 21
\][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[
y = 21 - 20 = 1
\][/tex]
5. Verify the solution with the second equation:
Using equation (2):
[tex]\[
3x - y = 14
\][/tex]
Substitute [tex]\( x = 5 \)[/tex] and [tex]\( y = 1 \)[/tex]:
[tex]\[
3(5) - 1 = 14
\][/tex]
[tex]\[
15 - 1 = 14
\][/tex]
Both sides are equal, confirming our solution is correct.
So, the solution to the system of equations is:
[tex]\[
\boxed{x = 5, y = 1}
\][/tex]
